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Steel Beam Design

  • #1
Hi.
I have started to do this question (the question it's self is attached with it's relevant diagrams) but I am coming up with a too large a maximum design load, the value I am looking for isn't on the tables I have so I must have made a calculation error somewhere. Any help would be much appreciated.
Here is what I have so far:
Volume of concrete slab= (6x6x0.175)=6.3m^3
Volume of reinforcement= 6.3 x 1.25% = 0.07875m^3
Total Volume= 6.37875m^3
G=(6.3x24.0) + (24.0 + 0.04725 x 0.07875)
=153.09kN/m^3
Q= 2.8 kn/m^3 given with the question

WL^*=(1.2G+1.5Q)x span x spacing
=(1.2x153.09+1.5x2.8)x6x3
=3382.34 kN
Which is a rather large number so I must be going wrong somewhere.
 

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  • #2
nvn
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mariechap89: You did not define the parameters you are using for your G calculation; and you did not include units in that calculation. Therefore, we do not really know what you are doing in your G calculation. Next time, you should define the parameters, and/or include units.

You have the wrong units on the live load. Read the given problem statement more carefully.

The volume of concrete per unit area is (1 m)(1 m)(0.175 m)/[(1 m)(1 m)] = 0.175 (m^3)/(m^2). Multiplying this by 24.0 kN/m^3 gives (24.0 kN/m^3)[0.175 (m^3)/(m^2)] = 4.200 kN/m^2. By the way, the specific weight of steel is gamma = 77.0 kN/m^3, not 24.0 kN/m^3.

Also, always leave a space between a numeric value and its following unit symbol. E.g., 6.3 m^3, not 6.3m^3. See the international standard for writing units (ISO 31-0).
 
  • #3
Thanks for your help, I realize I have got the calculation very wrong not even including the weight of the steel beam. I am pretty sure we are supposed to ignore the self weight of the beam for these calculations. I'm not sure how to use the weight of the concrete slab in the calculation for the steel beam sizing. Anyhelp would be great thanks.
 
  • #4
nvn
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mariechap89: The concrete dead load is in post 2. Finish that calculation by adding the reinforcement steel weight per unit area to it. After you obtain the factored dead plus live load per unit area, multiply it by spacing; i.e., multiply it by the width of the area supported by beam CD. This will give you a uniformly-distributed load, w, along the beam, in units of kN/m. Analyze the beam using the applied, uniformly-distributed load, w.
 
  • #5
Here is what I have now:
Volume of concrete= 6x6x0.175=6.3 m^3
Then the unit weight for concrete, dense aggregate, un-reinforced from a table is 24.0 kN/m^3 and then adding 0.6 for each 1% by volume of steel reinforcement the unit weight is 24.6015 kN/m^3.
Therefore G= 6.3x24.6015= 155 kN/m^2
I am given Strength limit state: WL=1.2G+1.5Q
Therefore WL=(1.2x155)+(1.5x2.8)x6x2= 2282.4 kN
Which, is still too large a value.

Also How would I handle the second part of the question ? I know how to handle a point load to find the steel beam size but how do I handle a point load and a udl ?

Thanks
 
  • #6
nvn
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See posts 2 and 4 to find out how to do this.

G = (24.0 kN/m^3)[(0.175 m^3)/(m^2)] + (77.0 kN/m^3)[0.0125(0.175 m^3)/(m^2)] = 4.200 kN/m^2 + 0.1684 kN/m^2 = 4.3684 kN/m^2.
Therefore, WL = 1.2*G + 1.5*Q = 1.2*(4.3684 kN/m^2) + 1.5*(2.8 kN/m^2) = 9.4421 kN/m^2.
Now multiply load by width supported by beam CD, to obtain the uniformly-distributed load, w, on beam CD.
w = WL*spacing = WL*(1.5 m + 1.5 m) = (9.4421 kN/m^2)(3 m) = 28.326 kN/m.

Or, if you want to use the table value (adding 0.6 kN/m^3 for each 1 % of steel), that would be as follows.

gamma = 24.0 kN/m^3 + (0.6 kN/m^3)*1.25 = 24.7500 kN/m^3.
G = (24.7500 kN/m^3)[(0.175 m^3)/(m^2)] = 4.3313 kN/m^2.
Therefore, WL = 1.2*G + 1.5*Q = 1.2*(4.3313 kN/m^2) + 1.5*(2.8 kN/m^2) = 9.3976 kN/m^2.
Now multiply load by width supported by beam CD, to obtain the uniformly-distributed load, w, on beam CD.
w = WL*spacing = (9.3976 kN/m^2)(3 m) = 28.193 kN/m.
 
  • #7
Hi.
Thanks for your help, this stuff is finally starting to make sense.
I understand how you've done the calcs for part a of the question and am pretty sure I know where I was going wrong.
I'm not too sure how to handle part b of the question though ??
 
  • #8
nvn
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You can compute the load supported by (coming down on) point D, which gives you just a point load at the midspan of beam MN.
 
  • #9
Hi
I have calculated the point loads on the girder MN (Ithink). I have used a 310UB40.4 beam. So 40.4 kg x 6 m = 242.4 kg
242.4 kg/2 = 121.1 kg (Asthe girder only supports half the weight of the beams)
121.1 x 9.81 = 1188.0 N = 1.188 kN
And there would be 3 point loads at 3m apart as girder MN supports 3 beams.
Assuming I have done this right, how ould I caluclate G from here including the udl from the concrete deck slab ?

Thanks
 
  • #10
nvn
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mariechap89: The factored UDL for beam CD is w in post 6. The factored point load at point D would be w*(0.5*LCD) + 1.2(1.189 kN), where LCD = length of beam CD = 6 m.

Your professor might let you ignore the point loads at points M and N, because they are virtually at the supports, causing virtually no moment on girder MN.
 

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